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Organic Chemistry · Acid-Base Theory

Organic Acids and Bases

Describe organic acid-base reactions using Brønsted-Lowry theory, identify the acid, base, conjugate acid, and conjugate base in each reaction, predict the direction of equilibrium from pKₐ values, and compare the relative acidity of organic compounds (O–H, C–H, and substituent-modified systems).

Theory — Acids, Bases, and pKₐ

The Brønsted-Lowry Definition

A Brønsted-Lowry acid is a proton (H⁺) donor. A Brønsted-Lowry base is a proton acceptor. Every acid-base reaction is a proton transfer between two species: the acid loses its proton and becomes a conjugate base; the base gains the proton and becomes a conjugate acid. The four species are always present together in an equilibrium:

Generic Brønsted-Lowry reaction H–A   +   :B   ⇌   :A⁻   +   H–B⁺
(acid)         (base)       (conjugate base)     (conjugate acid)
The acid and its conjugate base are called a conjugate acid-base pair — they differ by one proton.
acid (H donor) base (H acceptor) conjugate acid conjugate base

The pKₐ Scale

The acidity of a Brønsted acid HA is measured by its acid dissociation constant Kₐ, the equilibrium constant for HA ⇌ H⁺ + A⁻ in water. Because Kₐ spans many orders of magnitude, chemists use pKₐ = −log₁₀ Kₐ. Lower pKₐ means a stronger acid.

pKₐ interpretation pKₐ = −log₁₀(Kₐ)         Kₐ = [H⁺][A⁻] / [HA]
Strong acid: pKₐ < 0 (e.g. HCl pKₐ ≈ −7; H₂SO₄ pKₐ ≈ −10)
Moderate acid: pKₐ ≈ 4–5 (carboxylic acids)
Weak acid: pKₐ ≈ 10–16 (water, alcohols, phenol)
Very weak acid: pKₐ > 20 (alkynes, ketones α-H, amines, alkanes)
Stronger acid ⇒ more acidic H ⇒ lower pKₐ. Stronger base ⇒ its conjugate acid is higher pKₐ.

Predicting the Direction of an Acid-Base Equilibrium

A Brønsted reaction favours the side with the weaker acid (the one with the higher pKₐ). This is because the weaker acid has the stronger conjugate base — but the stronger conjugate base has already been deprotonated, so equilibrium settles on the side where both species are at their less-reactive (weaker) forms.

Quantitative rule Keq ≈ 10^(pKₐproduct-side acid − pKₐreactant-side acid)
If ΔpKₐ > 0 (product acid weaker) → equilibrium favours products (right).
If ΔpKₐ < 0 (product acid stronger) → equilibrium favours reactants (left).
A rule of thumb: if pKₐ of the conjugate acid on the product side exceeds the reactant acid's pKₐ by > 5 units, the reaction goes "essentially to completion."

Reference pKₐ Values

The following pKₐ values cover the most common O–H and C–H acids you will meet. Notice the range of pKₐ spans over 50 orders of magnitude — from very strong acids below 0 to very weak C–H acids near 50.

Acid (HA)pKₐType / Notes
H₂SO₄ −10 Inorganic strong acid
H₃O⁺ −1.7 Hydronium (reference for strong acids in water)
Trifluoroacetic acid (CF₃COOH) 0.23 3 × F EWG lowers pKₐ dramatically vs. acetic acid
Chloroacetic acid (ClCH₂COOH) 2.86 One Cl EWG
Formic acid (HCOOH) 3.75 Simplest carboxylic acid
Acetic acid (CH₃COOH) 4.76 Representative carboxylic acid
Benzoic acid (PhCOOH) 4.20 Aromatic COOH
2,4-Pentanedione (acac) 9.0 C-H between two C=O; doubly activated α-H
Ammonium (NH₄⁺) 9.25 Reference for amine basicity
para-Nitrophenol 7.15 Strong EWG on phenol dramatically increases acidity
Phenol (PhOH) 10.0 Aromatic OH — resonance stabilises PhO⁻
Bicarbonate (HCO₃⁻) 10.3 Reference buffer
Water (H₂O) 15.7 Reference for hydroxide base
Ethanol (CH₃CH₂OH) 16 Sp³ O–H
Acetaldehyde (α-H of CH₃CHO) 17 α-H to carbonyl
tert-Butanol 17 Steric/EDG raise pKₐ vs ethanol
Acetone (α-H) 19.3 α-H to a single C=O
Ethyl acetate (α-H) 25 α-H to ester carbonyl
Terminal alkyne (HC≡C–H) 25 Sp C–H — higher s-character = more acidic
Ammonia (NH₃) 38 N–H of amine (note: NH₄⁺ is the conjugate acid)
Ethylene (sp² C–H) 43 Sp² C–H
Methane / Ethane (sp³ C–H) ~50 Least acidic C–H

Factors That Influence Acidity

The acidity of an H–X bond (pKₐ of HA) depends on the stability of the conjugate base A⁻. Anything that stabilises the negative charge on A⁻ lowers the pKₐ. Four major factors:

1. Electronegativity of X

Within a row: more electronegative atoms stabilize negative charge better → lower pKₐ. H–F (pKₐ 3.2) < H–OH (15.7) < H–NH₂ (38) < H–CH₃ (~50).

2. Hybridization

Lone pair in hybrid orbital with more s-character is held closer to nucleus and is more stable. Sp C–H (pKₐ 25) < sp² C–H (43) < sp³ C–H (50).

3. Resonance stabilisation

Carboxylate (RCOO⁻) spreads the negative charge across 2 oxygens → carboxylic acids (pKₐ ~4–5) much more acidic than alcohols (pKₐ ~16). Similarly PhO⁻ stabilises the charge by resonance onto the ring, making phenol (pKₐ 10) more acidic than ethanol (pKₐ 16).

4. Inductive effects (EWG vs EDG)

Electron-withdrawing groups (F, Cl, NO₂, CF₃) stabilise conjugate base by pulling electron density through σ bonds → lower pKₐ. Electron-donating groups (CH₃, alkyl, OR) destabilise the conjugate base → higher pKₐ. Example: CF₃COOH (pKₐ 0.23) ≫ CH₃COOH (4.76).

α-Hydrogen Acidity of Carbonyl Compounds

Hydrogens on the carbon α to a carbonyl (C=O) are unusually acidic because the conjugate base — an enolate — is resonance-stabilised:

Enolate resonance CH₃–CO–CH₃ (pKₐ 19.3)   ⇌   CH₃–CO–CH₂⁻   ↔   CH₃–C(–O⁻)=CH₂
(α-H on acetone)       (C-centred enolate)   (O-centred enolate)
The negative charge is delocalised to the electronegative oxygen, stabilising the enolate and making α-H much more acidic than a typical sp³ C–H (50).

When the α-carbon sits between two carbonyls (e.g., 2,4-pentanedione or diethyl malonate), the enolate is stabilised by resonance with both oxygens — dropping the pKₐ further to 9–13.

Section I — Label and Predict

Six organic acid-base reactions. For each, identify the acid, base, conjugate acid, and conjugate base, then predict whether the equilibrium favours the products (right) or reactants (left) using pKₐ differences.

Section II — Rank by Acidity

Two rounds. Round 1 ranks O–H acids with different substituents (alcohols vs phenols vs carboxylic acids with EWGs). Round 2 ranks C–H acids (sp³ vs sp² vs sp, α-H, doubly-activated). Followed by four unknowns identified from pKₐ + structural clues.

Instructions — Running the Virtual Experiment

Section I — Label Components and Predict Direction

1
A reaction equation is displayed with 4 species (two reactants, two products). Select a role from the toolbar (acid, base, conjugate acid, or conjugate base), then click the species you want to give that role to. The species is colour-coded as you label it.
2
When all four species are labelled, examine the pKₐ panel that appears. It shows the pKₐ of the acid (left side) and the pKₐ of the conjugate acid (right side). Calculate ΔpKₐ mentally and pick whether the equilibrium favours products (right), reactants (left), or is approximately balanced.
3
Click Check Answer. If the labels are correct, the direction prediction is checked using the pKₐ rule (Keq ≈ 10^(pKₐ_product − pKₐ_reactant)). Click Next Reaction to proceed.
4
Work through all six reactions before moving to Section II. Examples span carboxylic acids, alcohols, phenols, terminal alkynes, amines, and α-H deprotonation of ketones.

Section II — Rank Acidity

1
Round 1 — O–H acids: six compounds (tert-butanol, ethanol, water, phenol, p-nitrophenol, CF₃COOH) appear as clickable cards. Click them in order from least acidic (highest pKₐ) to most acidic (lowest pKₐ). Your chosen order appears in the "Your sequence" strip below. Click Check Order to grade; the pKₐ appears on each card once it's in place.
2
Round 2 — C–H acids: same exercise with six C–H compounds of varying pKₐ (methane, ethylene, ammonia, terminal alkyne, acetone α-H, acetylacetone). Rank them least → most acidic.
3
Four unknowns: each unknown card gives a pKₐ value and a structural clue. Pick the compound that matches from the dropdown. Click Reveal Unknowns to check your answers.

Simulation — Acid-Base Virtual Bench

Organic Acids & Bases Lab | Section I — Label components and predict direction
Label each species with its Brønsted-Lowry role, then predict the direction of equilibrium.
1 of 6
Apply role:
Equilibrium favours:
Click the compounds in order, from least acidic (highest pKₐ) to most acidic (lowest pKₐ).
← LEAST acidic (high pKₐ) MOST acidic (low pKₐ) →
Compound pool (click in order):
Your sequence (least → most acidic):

Four Unknowns — Identify each acid from its pKₐ + structural clue

Each unknown provides its pKₐ and a structural description. Pick the compound from the dropdown.

Team Questions

Question 1. For the reaction CH₃COOH + NH₃ ⇌ CH₃COO⁻ + NH₄⁺, identify the acid, base, conjugate acid, and conjugate base. Given pKₐ(CH₃COOH) = 4.76 and pKₐ(NH₄⁺) = 9.25, does the equilibrium favour reactants or products?
Question 2. Rank ethanol (pKₐ 16), phenol (pKₐ 10), and acetic acid (pKₐ 4.76) in order of increasing acidity. Briefly explain why the ranking goes in this direction.
Question 3. Why is the α-hydrogen of acetone (pKₐ 19.3) so much more acidic than the C–H of ethane (pKₐ ~50) even though both are sp³ C–H bonds? Answer in one sentence referring to the conjugate base.
Question 4. Trifluoroacetic acid (CF₃COOH, pKₐ 0.23) is about 30,000× more acidic than acetic acid (CH₃COOH, pKₐ 4.76). Which electronic effect is responsible, and which direction do the C–F dipoles pull electron density?
Question 5. To deprotonate a terminal alkyne (pKₐ 25), chemists use sodium amide (NaNH₂). Explain why NaOH (conjugate base of water, pKₐ 15.7) cannot be used, but NaNH₂ (conjugate base of NH₃, pKₐ 38) can.

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.

Organic Acids and Bases

Chemistry 221 | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To identify the acid, base, conjugate acid, and conjugate base in representative organic Brønsted acid-base reactions; to use pKₐ values to predict the direction of equilibrium in each reaction; and to rank a set of organic O–H and C–H acids in order of increasing acidity based on structural features (electronegativity, hybridization, resonance, and inductive effects).

Theory

In Brønsted-Lowry theory, an acid donates a proton (H⁺) and a base accepts one. Every acid-base reaction is the transfer of a single proton between the two species, producing a conjugate acid-base pair on each side of the equilibrium: the acid (HA) becomes its conjugate base (A⁻), and the base (B) becomes its conjugate acid (BH⁺). The acid dissociation constant Kₐ (or its logarithmic form pKₐ = −log Kₐ) quantifies acid strength, with lower pKₐ corresponding to a stronger acid. An acid-base equilibrium favours the side with the weaker acid (higher pKₐ); quantitatively Keq ≈ 10^(pKₐproduct-side acid − pKₐreactant-side acid).

The acidity of an H–X bond depends on the stability of the conjugate base A⁻. Four effects control this stability: (i) electronegativity of the atom bearing the charge (O⁻ more stable than N⁻ more stable than C⁻, within the same row); (ii) hybridization (sp lone pair is closer to the nucleus than sp³, so sp C–H is most acidic of the C–H family); (iii) resonance stabilisation of the conjugate base (carboxylate spreads charge across two oxygens, phenolate spreads charge onto the ring, enolate spreads charge onto oxygen); and (iv) inductive effects of substituents (EWGs like F, Cl, NO₂ stabilise the conjugate base, lowering pKₐ; EDGs like alkyl groups destabilise it, raising pKₐ).

Calculations / Worked Examples

Sample reaction — CH₃COOH + NH₃ ⇌ CH₃COO⁻ + NH₄⁺:

Acid (reactants): CH₃COOH, pKₐ = 4.76
Base (reactants): NH₃
Conjugate base (products): CH₃COO⁻
Conjugate acid (products): NH₄⁺, pKₐ = 9.25
ΔpKₐ = 9.25 − 4.76 = +4.49
Keq ≈ 10^(+4.49) ≈ 3.1 × 10⁴
Equilibrium favours PRODUCTS (right) — acetic acid is the stronger acid, so it gives up its proton to NH₃ readily.

Sample ranking — O–H acids (increasing acidity):

tert-butanol (17) < ethanol (16) < water (15.7) < phenol (10) < p-nitrophenol (7.15) < CF₃COOH (0.23)
Reasoning: tert-butanol's three alkyl groups donate electron density inductively, destabilising its conjugate base (t-BuO⁻) and raising pKₐ above ethanol. Water is slightly more acidic than primary alcohols because the hydroxide conjugate base has less alkyl destabilisation. Phenol (pKₐ 10) is much more acidic than any aliphatic alcohol because the phenolate anion is resonance-stabilised by delocalisation onto the aromatic ring (3 additional resonance structures place negative charge on ortho/para carbons). Adding a para-nitro group to phenol further lowers pKₐ to 7.15: the nitro group is a strong EWG that stabilises the phenolate through both inductive pull and resonance with the ring. Trifluoroacetic acid's three fluorines pull electron density inductively from the carboxylate, stabilising it dramatically — CF₃COO⁻ is so stable that CF₃COOH has pKₐ ≈ 0, making it ~3 × 10⁴ times more acidic than acetic acid.

Results Table

Section I — component identification and equilibrium direction

ReactionAcidBaseConj. acidConj. baseΔpKₐFavours
CH₃COOH + NH₃ CH₃COOH (4.76)NH₃ NH₄⁺ (9.25) CH₃COO⁻+4.5Products
PhOH + HO⁻ PhOH (10.0) HO⁻ H₂O (15.7) PhO⁻ +5.7Products
HC≡CH + NaNH₂ HC≡CH (25) NH₂⁻ NH₃ (38) HC≡C⁻ +13 Products (strongly)
CH₃OH + NH₃ CH₃OH (15.5) NH₃ NH₄⁺ (9.25) CH₃O⁻ −6.3Reactants
Acetone + HO⁻ Acetone (19.3)HO⁻ H₂O (15.7) enolate−3.6Reactants
CH₃COOH + HC≡C⁻ CH₃COOH (4.76)HC≡C⁻ HC≡CH (25) CH₃COO⁻+20 Products (completely)

Section II — acidity rankings

RoundCompounds (least → most acidic)pKₐ values
Round 1 (O–H)t-BuOH < EtOH < H₂O < PhOH < p-NO₂-PhOH < CF₃COOH17, 16, 15.7, 10, 7.15, 0.23
Round 2 (C–H)CH₄ < CH₂=CH₂ < NH₃ < HC≡CH < acetone α-H < acac50, 43, 38, 25, 19.3, 9.0

Discussion

Section I confirmed that the direction of every Brønsted equilibrium is predicted by the pKₐ difference between the acid on each side. Reactions where the reactant acid has a lower pKₐ than the conjugate acid formed on the product side (positive ΔpKₐ) proceed forward — CH₃COOH is a stronger acid than NH₄⁺, so it transfers its proton to NH₃ almost completely (Keq ≈ 3 × 10⁴). Conversely, attempting to deprotonate a ketone α-H (pKₐ 19.3) with hydroxide (H₂O pKₐ 15.7) is unfavourable because the enolate is a stronger base than hydroxide — less than 1 in 10⁴ molecules is deprotonated at equilibrium, which is why kinetically-controlled enolate chemistry typically uses stronger bases like LDA (pKₐ of HN(i-Pr)₂ ≈ 36) rather than NaOH.

The α-H acidity of carbonyls is a clear illustration of resonance stabilisation: although acetone's α-C–H is formally an sp³ C–H bond that would normally be pKₐ ~50, the resulting enolate delocalises the negative charge onto the electronegative carbonyl oxygen, dropping the pKₐ to 19.3 — a stabilisation of over 30 pKₐ units (10³⁰ in equilibrium constant). Adding a second carbonyl (as in 2,4-pentanedione or diethyl malonate) delocalises the charge onto two oxygens, dropping pKₐ further to 9.0, comparable to phenol and ammonium.

Section II's rankings exposed the four factors driving acidity in a quantitative way. Round 1 (O–H acids) spans 17 pKₐ units (10¹⁷-fold difference in Kₐ), driven almost entirely by conjugate-base stabilisation. Aliphatic alkoxides sit at the top (least acidic) because alkyl groups destabilise the negative oxygen through weak electron donation. Phenoxide benefits from resonance into the ring, dropping pKₐ to 10; a p-nitro group reinforces this via both resonance and induction, reaching 7.15. Carboxylate's two-oxygen resonance drops acetic acid to 4.76, and triple-fluorine inductive stabilisation on CF₃COO⁻ reaches pKₐ ≈ 0. Round 2 (C–H acids) emphasised hybridization (sp 25 < sp² 43 < sp³ 50) and the dramatic effect of α-conjugation (acetone 19.3; acac 9.0).

Conclusion

All six reactions in Section I were labelled correctly and the directions predicted quantitatively via ΔpKₐ; four of six favoured products, two favoured reactants, matching the rule that equilibrium favours the weaker acid. The O–H and C–H acidity rankings in Section II reproduced the expected trends based on electronegativity, hybridization, resonance, and inductive effects, with the full range of organic acidity spanning nearly 50 pKₐ units. All four unknowns were identified from their pKₐ and structural clues. The experiment confirmed that conjugate-base stability, not intrinsic proton "affinity", is the organising principle of organic acid-base chemistry.

Practice Questions

Show your reasoning. Use pKₐ values from the reference table above where relevant.

Question 1
For each reaction, identify the acid, base, conjugate acid, and conjugate base: (a) HCOOH + H₂O ⇌ HCOO⁻ + H₃O⁺; (b) CH₃NH₂ + HCl ⇌ CH₃NH₃⁺ + Cl⁻.
Hint: identify the proton donor and acceptor on each side. The species that LOST a proton going right-to-left is the conjugate acid on the right.
Question 2
Rank the following from least to most acidic and justify: methanol (pKₐ 15.5), 2-chloroethanol (pKₐ 14.3), 2,2-dichloroethanol (pKₐ 12.2), 2,2,2-trichloroethanol (pKₐ 12.2). How does halogen substitution change the pKₐ?
Hint: each Cl is an EWG that pulls electron density toward itself. The effect is strongest when Cl is closest to the conjugate-base oxygen.
Question 3
Predict the direction of equilibrium: CH₃OH + NaH ⇌ CH₃O⁻Na⁺ + H₂. Given that H₂ has pKₐ ≈ 35 and methanol has pKₐ ≈ 15.5, estimate Keq and briefly explain why NaH is a common reagent for deprotonating alcohols.
Hint: ΔpKₐ ≈ 35 − 15.5 = +20, so Keq ≈ 10²⁰ — reaction goes to completion. NaH releases H₂ gas which also drives the reaction forward.
Question 4
Draw the two resonance structures of the acetate anion (CH₃COO⁻). How does this help explain why acetic acid (pKₐ 4.76) is ~10¹¹ times more acidic than ethanol (pKₐ 16)?
Hint: acetate has the negative charge delocalised equally over two oxygens (equivalent resonance forms). Ethoxide localises the charge on a single oxygen.
Question 5
Why is the α-hydrogen of cyclohexanone more acidic (pKₐ ≈ 20) than the β-hydrogen (pKₐ ≈ 45)? Draw the conjugate base formed by α-deprotonation and show the enolate resonance structures.
Hint: α-deprotonation gives an enolate with resonance into C=O. β-deprotonation gives a carbanion with no resonance stabilisation.
Question 6 — Challenge
The pKₐ of the C–H bond in chloroform (HCCl₃) is about 25, similar to that of a terminal alkyne. Given that chloroform is sp³ while the alkyne is sp, explain how such different-hybridised C–H bonds can have the same pKₐ. What stabilising factor is at play in HCCl₃?
Hint: three chlorines are very strong EWGs. They stabilise the CCl₃⁻ conjugate base inductively, compensating for the lower s-character of sp³ C–H.